Question: The equation of a circle $C$ is $x^2+y^2+16x+6y+64 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+16x) + (y^2+6y) = -64$ $(x^2+16x+64) + (y^2+6y+9) = -64 + 64 + 9$ $(x+8)^{2} + (y+3)^{2} = 9 = 3^2$ Thus, $(h, k) = (-8, -3)$ and $r = 3$.